A chemist has two solutions of HNO_3. One has a 40% concentration and the other has a 25% concentration. How many liters of each solution must be mixed to obtain 120 L of a 30% solution?

Let x be the amount of the 40% solution needed and y be the amount of the 25% solution needed. We know that the total volume of the mixture is 120 L and the concentration of the mixture is 30%. This means that: 0.4x + 0.25y = 0.3(120) Simplifying this equation, we get: 0.4x + 0.25y = 36 We also know that the total amount of the two solutions is 120 L, so: x + y = 120 We now have two equations with two unknowns. We can solve for x and y by using substitution or elimination methods. Here, we will use the substitution method: x = 120 - y Substituting this into the first equation, we get: 0.4(120 - y) + 0.25y = 36 48 - 0.4y + 0.25y = 36 0.15y = 12 y = 80 So, the chemist needs 80 L of the 25% solution and: x = 120 - y = 120 - 80 = 40 40 L of the 40% solution to obtain 120 L of a 30% solution.